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Chapter 8: Problem 76

Would you expect AlN to have a lattice energy that is larger or smaller than ScN? Explain.

Short Answer

Expert verified
The lattice energy of Aluminum Nitride (AlN) is expected to be larger than that of Scandium Nitride (ScN). This is because both compounds have the same magnitude of charge on their metal and non-metal ions, but AlN has smaller metal ions (Al鲁鈦�) compared to ScN (Sc鲁鈦�). Smaller ions result in a stronger electrostatic interaction, leading to a larger lattice energy in AlN.

Step by step solution

01

Compare the charges of the ions of AlN and ScN

Aluminum forms a 3+ ion (Al鲁鈦�) when it bonds with the Nitride ion (N鲁鈦�) to form AlN, while Scandium forms a 3+ ion (Sc鲁鈦�) when it bonds with the Nitride ion (N鲁鈦�) to form ScN. Thus, both compounds have the same magnitude of charge on their metal and non-metal ions.
02

Compare the sizes of the ions of AlN and ScN

In order to compare the sizes of the ions in each compound, we can look at their ionic radii. The ionic radius of Aluminum (Al鲁鈦�) is 50 pm, while the ionic radius of Scandium (Sc鲁鈦�) is 75 pm. The ionic radius of Nitride ion (N鲁鈦�) is the same in both compounds - 146 pm.
03

Determine the difference in lattice energy

Since both AlN and ScN have the same magnitude of charge on their metal and non-metal ions, the difference in lattice energy depends on the size of the ions. AlN has smaller metal ions (Al鲁鈦�) compared to ScN (Sc鲁鈦�). As smaller ions result in a stronger electrostatic interaction, the lattice energy of AlN is expected to be larger than that of ScN.
04

Conclusion

The lattice energy of Aluminum Nitride (AlN) is expected to be larger than that of Scandium Nitride (ScN) due to the smaller size of the Aluminum ion (Al鲁鈦�) compared to the Scandium ion (Sc鲁鈦�), which results in a stronger electrostatic interaction between the ions in AlN.

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Most popular questions from this chapter

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

Use Lewissymbols and Lewis structures to diagram the formation of \(\mathrm{PF}_{3}\) from \(\mathrm{P}\) and \(\mathrm{F}\) atoms.

For each of the following molecules or ions of sulfur and oxygen, write a single Lewis structure that obeys the octet rule, and calculate the oxidation numbers and formal charges on all the atoms: (a) \(\mathrm{SO}_{2}\), (b) \(\mathrm{SO}_{3}\), (c) \(\mathrm{SO}_{3}{ }^{2-}\). (d) Arrange these molecules/ions in order of increasing \(\mathrm{S}-\mathrm{O}\) bond distance.

(a) What are valence electrons? (b) How many valence electrons does a nitrogen atom possess? (c) An atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} .\) How many valence electrons does the atom have?

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) \(\mathrm{NO}^{+}\), (b) \(\mathrm{POCl}_{3}\) (P is bonded to the three \(\mathrm{Cl}\) atoms and to the \(\mathrm{O}\) ), (c) \(\mathrm{ClO}_{4}^{-}\), (d) \(\mathrm{HClO}_{3}(\mathrm{H}\) is bonded to \(\mathrm{O})\).
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